8节点矩形单元一致质量矩阵推导？

shipsj

18# pasuka
不知 库克，《有限元分析的概念和应用》此书和版次说详细上了这个问题，一是还没弄清，自己先积一下，看看

pasuka

18# pasuka
这个结果和高斯积分点有关
我提的问题还没有用高斯积分的方法去实现，不过你的资料和数据好像和Zinkiewicz的书上不一样了，我去找下这边书
shipsj 发表于 2010-7-3 16:59

lz的脑筋还是没有转过来
3*3点高斯积分点结果和matlab的int符号积分点结果一致，都是精确积分！！！
王勖成书上是2*2点高斯积分结果

shipsj

18# pasuka
不知 库克，《有限元分析的概念和应用》此书何版次上详细说了这个问题，一时还没不知，自己先积一下，看看

shipsj

22# pasuka
可是王的书上却说是用缩放因子的方法来的啊，Zinkiewicz的书上也是这么说的，所以我要积一下，看看结果

shipsj

clear
syms x y cofx cofe;
%corner node 1,2,3,4 shape function
Ni=1/4*(1+cofx*x)*(1+cofe*y)*(cofx*x+cofe*y-1);
N1=subs(Ni,{cofx,cofe},{-1,-1});
N2=subs(Ni,{cofx,cofe},{1,-1});
N3=subs(Ni,{cofx,cofe},{1,1});
N4=subs(Ni,{cofx,cofe},{-1,1});
%middle node 5,6,7,8 shape function
N5=1/2*(1-x^2)*(1-y);
N6=1/2*(1-y^2)*(1+x);
N7=1/2*(1-x^2)*(1+y);
N8=1/2*(1-y^2)*(1-x);
%Nsh is composed by shape functions
%Nsh=[0 N1 0 N2 0 N3 0 N4 0 N5 0 N6 0 N7 0 N8;N1 0 N2 0 N3 0 N4 0 N5 0 N6 0 N7 0 N8 0]
%We can simply vertify through u or v degree of freedom
Nsh=[N1 N2 N3 N4 N5 N6 N7 N8];
Masstr=Nsh.' * Nsh;
ngas=2;                                       
posg=sym(zeros(1,ngas));
weig=sym(zeros(1,ngas));
if(ngas==2)
        posg(1)=-1/sqrt(3);
        weig(1)=1;
%nags=3,just thinking this two suitation
else
        posg(1:2)=[-sqrt(3/5) 0];
        weig(1:2)=[5/9 8/9];
end
kgas=floor(ngas./2);
for ig=1:kgas
        jg=ngas+1-ig;
        posg(jg)=-posg(ig);
        weig(jg)=weig(ig);
end
Massst=sym(zeros(length(Masstr)));
Massco=Massst;
for ig=1:ngas
        for jg=1:ngas
                Massst=subs(Masstr,{x,y},{posg(ig),posg(jg)});
                Massco = Massco + weig(ig) .* weig(jg) .* Massst;
        end
end
Massco=simplify(Massco);               
vs=sym(zeros(1,length(Massco)));
for i=1:length(Massco), vs(i) = Massco(i,i); end
sum(vs) .\ vs
经pasuka的指点，用2*2节点高斯积分得出结果[ 1/36, 1/36, 1/36, 1/36,  2/9,  2/9,  2/9,  2/9]
哎，王勖成的书上在介绍协调质量矩阵对角化时压根就没提高斯积分的这种方法，Zinkiewicz的书上倒是说了，却白纸黑字的写着是用缩放因子的方法推来的，这个是否可以向出版社反应一下，做个勘误啊，王勖成的书，在动力学这块我还发现了另外的一个公式有问题，有时间写出来，以免大家再费时间

pasuka

clear
syms x y cofx cofe;
%corner node 1,2,3,4 shape function
Ni=1/4*(1+cofx*x)*(1+cofe*y)*(cofx*x+cofe*y-1);
N1=subs(Ni,{cofx,cofe},{-1,-1});
N2=subs(Ni,{cofx,cofe},{1,-1});
N3=subs(Ni,{cofx,co ...
shipsj 发表于 2010-7-3 18:59

用得都是缩放因子的方法，但是2点高斯积分和3点高斯积分得到的质量阵不同，结果自然不一样
此外，对于中厚板单元，一般会采用缩减积分，而不是完全积分

shipsj

恩，计算的过程都用了缩放法，不过现在我手头的书本，缩放法指的都是我前面的推导方法。感谢各位朋友的参与，终于让这个问题可以告一段落！谢谢各位！

shipsj

这是我目前发现的错误，应这版有人问起，特在此版说明
http://forum.simwe.com/thread-935234-1-1.html

caoer

Well. First of all, I think Adina is a very good software. I don't know what kind of coding language Adina is using. As you know, Adina is originally from NONSAP written in Fortran.

As far as I kno ...
tonnyw 发表于 2010-7-3 01:21

Let the expert answer you .

1. 
   
2. Lagrangian Vs. Serendipity Finite Elements    Submitted by elabbasi on Sat, 2010-06-05 03:29.
   
3.     Dear Jayadeep,
   
4. 
   
5. Both the 2D 8-node Serendipity and 9-node Lagrangian elements can exactly represent up to quadratic displacement fields if the elements are  square or rectangular, as stated in Bathe's Finite Element Procedures book. If they only have angular distortion there is a difference between the two elements. The 9-node Lagrangian can still exactly represent a quadratic displacement field while the 8-node Serendipity element cannot. If the elements are distorted beyond just angular distortion (curved sides for example) neither of them can exactly represent quadratic displacements. See Fig. 5.17 in Bathe's book for an example.
   
6. 
   
7. Another advantage for the 9-node element over the 8-node one is in its mixed "u/p" formulation, also explained in Bathe's book (see Table 4.6). The 9-node element satisfies the relevant inf-sup condition with 3 pressure degrees of freedom while the 8-node element does not. In that case, either 1 or 3 pressure DOF can be used and while both choices work in many cases, neither are mathematically ideal. The same arguments apply to the 3D 20-node Serendipity and the 27-node Lagrangian elements (this time with 4 pressure DOFs).
   
8. 
   
9. Another advantage that only applies to the 3D version of these elements is in contact. Enforcing contact constraints at the nodes (which most commercial codes do) is tricky with the 20-node element. That is because some of the nodal contact forces will be tensile even though the contact tractions are compressive! You can check that out by applying a constant pressure to a 20-node element face and calculating the nodal forces. The corner nodal forces will be in the opposite direction to the pressure. 27-node elements do not have that problem. There are some workaround to that issue though, so it's not as bad as it looks.
   
10.   
    
11. Regarding commercial codes, the Lagrangian elements are available in Abaqus, Adina, and NX Nastran, and, as far as I know, not available in Ansys or Marc.
    
12. 
    
13. I think the main reason some commercial codes do not adopt Lagrangian elements is cost (positioning the extra nodes should not be a problem). In 2D the ratio of nodes in a uniform fine mesh of 8-node and 9-node elements is 3:4. However in 3D it's 1:2. Also, the bandwidth goes up not just the number of degrees of freedom. The increase in computational time can be significant.
    
14. Regards,
    
15. Nagi
    
16. 
    

复制代码

From http://imechanica.org/node/8325

tonnyw

29# caoer

Thanks, this is something I didn't know. I also located a paper talking about this.

tonnyw

25# shipsj
This is strange. I didn't see anywhere in the book saying that reduced integration is used.

We are using quadratic element and the highest polynomial order of mass matrix is 4. To integrate exactly, at least 3X3 integration points should be used.

shipsj

31# tonnyw
对此我也很感到很奇怪，不知库克那本书有没有解释

tonnyw

32# shipsj

Here is the part from Cook's book.

zh123168

还没看这本书，看过的都把错误说说吧，以免别人再走弯路，呵呵！

wjd1984

我也遇到这个问题，同求，谢谢